\documentclass[spanish,a4paper,11pt]{article}

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\begin{document}
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\input{parcialito1_tema1.tex}

\paragraph{1)a)} Para que resulte $\mathbf{A} \perp \mathbf{B}$ debe ser que $\mathbf{A}\cdot\mathbf{B}=0$. O sea:
\begin{align*}
\mathbf{A}\cdot\mathbf{B} &=0 \\
(2\,\mathbf{I}-3\,\mathbf{J}+\mathbf{K})\cdot(k\,\mathbf{I}-\mathbf{K}) &=0
\end{align*}
Y ahora el producto escalar de dos vectores (que da como resultado un n'umero) se hac'ia x por x, m'as, y por y, m'as, z por z:
\begin{align*}
(2\,\mathbf{I}-3\,\mathbf{J}+\mathbf{K})\cdot(k\,\mathbf{I}-\mathbf{K}) &=0 \\
2k+(-3)0+1(-1) &=0\\
2k-1 &=0\\
k &=\dfrac{1}{2}
\end{align*}

\emph{Respuesta:} \fbox{$k=\dfrac{1}{2}$}

\paragraph{1)b)} $\mathsf{Proy}_\mathbf{B}(\mathbf{A})= \dfrac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{B}|^2}\,\mathbf{B}$\\

\noindent
En el punto 1)a) averiguamos que debe ser $k=\frac{1}{2}$ para que $\mathbf{A}\cdot\mathbf{B}=0$. O sea que:$$\dfrac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{B}|^2} = \dfrac{0}{|\mathbf{B}|^2} = 0$$ Es decir que: $$\mathsf{Proy}_\mathbf{B}(\mathbf{A})= \dfrac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{B}|^2}\,\mathbf{B} = 0\,(k\,\mathbf{I}-\mathbf{K}) = 0\,\left(\tfrac{1}{2},0,-1\right) = (0,0,0)$$

\emph{Respuesta:} \fbox{$\mathsf{Proy}_\mathbf{B}(\mathbf{A})= (0,0,0)$}

\paragraph{2)a)} \emph{Respuesta:} 
$\boxed{
\left\{
\begin{aligned}
x &= t+1 \\ 
y &= t^2+1 
\end{aligned} 
\right. \quad \text{con} \quad -1\leq t\leq 3
}
$	

\paragraph{2)b)} 
De $x=t+1$ tenemos que $x-1=t$\\

Luego, si $y=t^2+1$ entonces $y=(\underbrace{x-1}_t)^2+1$. O sea que $y=(x-1)^2+1$ \\

Como $-1\leq t\leq 3$ entonces $-1\leq \underbrace{x-1}_{t} \leq 3$, o sea que, sumando $1$ en todos los miembros de la inecuaci'on, obtenemos que $0\leq x\leq 4$\\

Entonces la ecuaci'on de la \emph{trayectoria} nos queda: $$y=(x-1)^2+1 \qquad 0\leq x \leq 4$$ Para graficarla observemos que est'a escrita en la forma $y=a(x-x_v)^2+y_v$, o sea que: $$a=1 \qquad x_v=1 \qquad y_v=1$$ Entonces ya sabemos que el v'ertice est'a en el $(1;1)$ y que es c'oncava hacia arriba:\\

\hspace{3cm}
%\begin{tikzpicture}[domain=0:2]
\begin{tikzpicture}[scale=0.5]
%\draw[thick,color=gray,step=.5cm,dashed] (2,0) grid (6,12);
\draw[->] (-4,0) -- (6,0) node[right] {$X$};
\draw[->] (0,-1) -- (0,18) node[above] {$Y$};
%\foreach \x/\xtext in {-2/-2, -1/-1, 1/1, 2/2, 3/3, 4/4, 5/5, 6/6}
%    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
%\foreach \y/\ytext in {1/1, 2/2, 2.25/2\frac{1}{4}, 3/3}
%    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
\foreach \x/\xtext in {1/1}
    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {1/1}
    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
%\draw plot[id=x] function{x*x};
\draw (-3,17) parabola bend (1,1) (5,17) node[below right] {$y=(x-1)^2+1$};
\end{tikzpicture}

\vspace{.5cm}
Ahora tengamos en cuenta que $0\leq x\leq 4$:
$$
\begin{array}{c|c}
x & y=(x-1)^2+1 \\
\hline
0 & 2 \\
4 & 10
\end{array}
$$ Y entonces el gr'afico nos queda desde el punto $(0;2)$ hasta el puntp $(4;10)$:\\

\hspace{3cm}
%\begin{tikzpicture}[domain=0:2]
\begin{tikzpicture}[scale=0.5]
%\draw[thick,color=gray,step=.5cm,dashed] (2,0) grid (6,12);
\draw[->] (-1,0) -- (5,0) node[right] {$X$};
\draw[->] (0,-1) -- (0,11) node[above] {$Y$};
%\foreach \x/\xtext in {-2/-2, -1/-1, 1/1, 2/2, 3/3, 4/4, 5/5, 6/6}
%    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
%\foreach \y/\ytext in {1/1, 2/2, 2.25/2\frac{1}{4}, 3/3}
%    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
\foreach \x/\xtext in {1/1, 4/4}
    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {2/2, 1/1, 10/10}
    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
%\draw plot[id=x] function{x*x};
\draw (0,2) parabola bend (1,1) (4,10) node[below right] {$y=(x-1)^2+1\quad \text{con} \quad 0\leq x\leq 4$};
\end{tikzpicture}

\vspace{.5cm}
\emph{Respuesta:} $\boxed{y=(x-1)^2+1 \qquad 0\leq x \leq 4}$

\fbox{
\hspace{3cm}
%\begin{tikzpicture}[domain=0:2]
\begin{tikzpicture}[scale=0.5]
%\draw[thick,color=gray,step=.5cm,dashed] (2,0) grid (6,12);
\draw[->] (-1,0) -- (5,0) node[right] {$X$};
\draw[->] (0,-1) -- (0,11) node[above] {$Y$};
%\foreach \x/\xtext in {-2/-2, -1/-1, 1/1, 2/2, 3/3, 4/4, 5/5, 6/6}
%    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
%\foreach \y/\ytext in {1/1, 2/2, 2.25/2\frac{1}{4}, 3/3}
%    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
\foreach \x/\xtext in {1/1, 4/4}
    \draw[shift={(\x,0)}] (0pt,2pt) -- (0pt,-2pt) node[below] {$\xtext$};
\foreach \y/\ytext in {2/2, 1/1, 10/10}
    \draw[shift={(0,\y)}] (2pt,0pt) -- (-2pt,0pt) node[left] {$\ytext$};
%\draw plot[id=x] function{x*x};
\draw (0,2) parabola bend (1,1) (4,10) node[below right] {$y=(x-1)^2+1 \quad \text{con} \quad 0\leq x\leq 4$};
\end{tikzpicture}
}

\end{document}
